Questions and Answers Type: | MCQ (Multiple Choice Questions). |

Main Topic: | Quantitative Aptitude. |

Quantitative Aptitude Sub-topic: | Probability Aptitude Questions and Answers. |

Number of Questions: | 10 Questions with Solutions. |

- If two dice are thrown, then find the probability of getting the total of \(10\)?
- \(\frac{1}{15}\)
- \(\frac{1}{13}\)
- \(\frac{1}{14}\)
- \(\frac{1}{12}\)

Answer: (d) \(\frac{1}{12}\)

Solution: number of possible outcomes = \(6 \times 6\) = \(36\)

number of favourable events = \(\{(4, \ 6), \ (5, \ 5), \ (6, \ 4)\}\)

number of favourable outcomes = \(3\), then $$ Probability = \frac{3}{36} $$ $$ = \frac{1}{12} $$

Solution: number of possible outcomes = \(6 \times 6\) = \(36\)

number of favourable events = \(\{(4, \ 6), \ (5, \ 5), \ (6, \ 4)\}\)

number of favourable outcomes = \(3\), then $$ Probability = \frac{3}{36} $$ $$ = \frac{1}{12} $$

- Two balls are to be drawn from a box containing \(10\) blue and \(20\) red balls. Find the probability of getting that one ball is blue and one ball is red?
- \(\frac{45}{32}\)
- \(\frac{40}{87}\)
- \(\frac{36}{57}\)
- \(\frac{40}{97}\)

Answer: (b) \(\frac{40}{87}\)

Solution: number of blue balls = \(10\)

number of red balls = \(20\), then $$ Probability = \frac{^{10}C_1 \times ^{20}C_1}{^{30}C_2} $$ $$ = \frac{10 \times 20}{435} = \frac{40}{87} $$

Solution: number of blue balls = \(10\)

number of red balls = \(20\), then $$ Probability = \frac{^{10}C_1 \times ^{20}C_1}{^{30}C_2} $$ $$ = \frac{10 \times 20}{435} = \frac{40}{87} $$

- If two coins are tossed, then find the probability of getting atleast one tail?
- \(\frac{3}{4}\)
- \(\frac{4}{3}\)
- \(\frac{2}{3}\)
- \(\frac{3}{2}\)

Answer: (a) \(\frac{3}{4}\)

Solution: sample space = \(\{HH, \ HT, \ TH, \ TT\}\)

number of possible outcomes = \(4\)

number of favourable outcomes = \(3\), then $$ Probability = \frac{3}{4} $$

Solution: sample space = \(\{HH, \ HT, \ TH, \ TT\}\)

number of possible outcomes = \(4\)

number of favourable outcomes = \(3\), then $$ Probability = \frac{3}{4} $$

- From a pack of \(52\) playing cards, three cards are drawn at random. what is the chance of getting two king and one queen?
- \(\frac{3}{5225}\)
- \(\frac{3}{2552}\)
- \(\frac{3}{5525}\)
- \(\frac{3}{5552}\)

Answer: (c) \(\frac{3}{5525}\)

Solution: total number of cards = \(52\)

number of cards of king = \(4\)

number of cards of queen = \(4\), then $$ Probability = \frac{^4C_2 \times ^4C_1}{^{52}C_3} $$ $$ = \frac{6 \times 4}{44200} $$ $$ = \frac{3}{5525} $$

Solution: total number of cards = \(52\)

number of cards of king = \(4\)

number of cards of queen = \(4\), then $$ Probability = \frac{^4C_2 \times ^4C_1}{^{52}C_3} $$ $$ = \frac{6 \times 4}{44200} $$ $$ = \frac{3}{5525} $$

- A bag contains \(7\) pink and \(9\) green balls, if four balls are picked at random then find the probability of getting \(2\) pink and \(2\) green balls?
- \(\frac{9}{120}\)
- \(\frac{5}{121}\)
- \(\frac{9}{130}\)
- \(\frac{6}{113}\)

Answer: (c) \(\frac{9}{130}\)

Solution: number of pink balls = \(7\)

number of green balls = \(9\), then $$ Probability = \frac{^7C_2 \times ^9C_2}{^{16}C_4} $$ $$ = \frac{21 \times 36}{10920} = \frac{9}{130} $$

Solution: number of pink balls = \(7\)

number of green balls = \(9\), then $$ Probability = \frac{^7C_2 \times ^9C_2}{^{16}C_4} $$ $$ = \frac{21 \times 36}{10920} = \frac{9}{130} $$

- A container contains \(20\) red and \(30\) blue balls. Find the probability of getting one red ball?
- \(\frac{2}{3}\)
- \(\frac{3}{2}\)
- \(\frac{4}{5}\)
- \(\frac{2}{5}\)

Answer: (d) \(\frac{2}{5}\)

Solution: number of red balls = \(20\)

number of blue balls = \(30\), then $$ Probability = \frac{^{20}C_1}{^{50}C_1} $$ $$ = \frac{20}{50} = \frac{2}{5} $$

Solution: number of red balls = \(20\)

number of blue balls = \(30\), then $$ Probability = \frac{^{20}C_1}{^{50}C_1} $$ $$ = \frac{20}{50} = \frac{2}{5} $$

- A box contains three red, six pink and seven black balls. If three balls are picked at random, then find the probability of getting three different colour balls?
- \(\frac{7}{60}\)
- \(\frac{9}{80}\)
- \(\frac{7}{80}\)
- \(\frac{9}{70}\)

Answer: (b) \(\frac{9}{80}\)

Solution: number of red colour balls = \(3\)

number of pink colour balls = \(6\)

number of black colour balls = \(7\)

then probability of getting three different colour balls, $$ Probability = \frac{^3C_1 \times ^6C_1 \times ^7C_1}{^{16}C_3} $$ $$ = \frac{3 \times 6 \times 7}{1120} $$ $$ = \frac{9}{80} $$

Solution: number of red colour balls = \(3\)

number of pink colour balls = \(6\)

number of black colour balls = \(7\)

then probability of getting three different colour balls, $$ Probability = \frac{^3C_1 \times ^6C_1 \times ^7C_1}{^{16}C_3} $$ $$ = \frac{3 \times 6 \times 7}{1120} $$ $$ = \frac{9}{80} $$

- There are two boxes, first one contains \(10\) black, \(20\) white balls and second one contains \(5\) black, \(7\) white balls. If one ball is drawn from each box, then find the probability of getting two black balls?
- \(\frac{5}{36}\)
- \(\frac{5}{32}\)
- \(\frac{7}{36}\)
- \(\frac{7}{32}\)

Answer: (a) \(\frac{5}{36}\)

Solution: first box contains \(10\) black and \(20\) white balls and

second box contains \(5\) black and \(7\) white balls

then probability of getting two black colour balls, $$ P = \frac{10}{30} \times \frac{5}{12}$$ $$ = \frac{5}{36} $$

Solution: first box contains \(10\) black and \(20\) white balls and

second box contains \(5\) black and \(7\) white balls

then probability of getting two black colour balls, $$ P = \frac{10}{30} \times \frac{5}{12}$$ $$ = \frac{5}{36} $$

- If a fair dice is thrown, then find the probability of getting a number greater than one?
- \(\frac{4}{5}\)
- \(\frac{5}{6}\)
- \(\frac{3}{7}\)
- \(\frac{4}{7}\)

Answer: (b) \(\frac{5}{6}\)

Solution: number of possible outcomes = \(6\)

number of favourable events = \(\{2, \ 3, \ 4, \ 5, \ 6\}\)

number of favourable outcomes = \(5\), then $$ Probability = \frac{5}{6} $$

Solution: number of possible outcomes = \(6\)

number of favourable events = \(\{2, \ 3, \ 4, \ 5, \ 6\}\)

number of favourable outcomes = \(5\), then $$ Probability = \frac{5}{6} $$

- From a pack of \(52\) playing cards, two cards are drawn at random. What is the probability of drawing a card of red colour?
- \(\frac{1}{2}\)
- \(\frac{1}{3}\)
- \(\frac{1}{4}\)
- \(\frac{1}{5}\)

Answer: (a) \(\frac{1}{2}\)

Solution: total number of cards = \(52\)

total number of red colour cards = \(26\), then $$ Probability = \frac{26}{52} $$ $$ = \frac{1}{2} $$

Solution: total number of cards = \(52\)

total number of red colour cards = \(26\), then $$ Probability = \frac{26}{52} $$ $$ = \frac{1}{2} $$

Lec 1: Introduction to Probability
Exercise-1
Lec 2: Addition Rule
Exercise-2
Lec 3: Multiplication Rule
Exercise-3
Exercise-4
Exercise-5